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Oksi-84 [34.3K]
2 years ago
7

Which of the following is an example of a physical change?(1 point)

Chemistry
2 answers:
WINSTONCH [101]2 years ago
8 0
Dissolving sugar in water is a example of physical change
user100 [1]2 years ago
4 0
Dissolving sugar in water
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You are measuring the mass of different chemicals to get ready to conduct an experiment.
Leto [7]

Answer:

Explanation:

B

4 0
2 years ago
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Which of the following statements is true about catalysts? and Why?
LUCKY_DIMON [66]
I believe the correct answer from the choices listed above is option D. Catalysts lower the activation energy of a chemical reaction. It <span>is a substance which speeds up a reaction, but is chemically unchanged at the end of the reaction. It provides another pathway for the reaction to occur.</span>
8 0
3 years ago
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When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
2 years ago
Develop a demonstration to show how mass is not the same thing as weight
pychu [463]

This is more of a physics explanation, but here we go.

Mass is a measure of how much "matter" is in an object. Weight is the force applied onto an object by gravity. Weight itself can be related to mass like this:

f_g = mg

where g is a gravitational constant. For our purposes, it's defined by whatever planet you are on. Following this, we can demonstrate that mass is NOT the same thing as weight if we take two objects of the same mass and put them on different planets.

Let E refer to Earth and F refer to Mars

g_E = 9.81 m/s^2\\g_F = 3.711 m/s^2

Following this, we can see clearly that weight is not the same as mass:

m*9.81 : m*3.711 \\\int\limits^a_b {x} \, dx 9.81 \neq 3.711\\f_g E \neq f_g F\\

If weight was the same thing as mass, the two values would be the same, as the mass of the two objects is the same. But since weight is defined in the context of gravity, they are not.

4 0
3 years ago
A 150.0 mL sample of a 1.50 M solution of CuSO4 is mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The t
svp [43]
Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225 
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
<span>specific heat solns = specific heat H2O = 4.18 J/K*C </span>

<span>CuSO4 + 2KOH = Cu(OH)2 + 2H2O </span>
<span>q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T </span>
<span>q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) </span>
<span>dHrxn in J/mol= q/0.225 mol CuSO4 </span>
<span>Then convert to kJ/mol


</span>
5 0
3 years ago
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